interpret Parigot's lambda-mu calculus in Haskell

Question

One can interpret the lambda calculus in Haskell:

data Expr = Var String | Lam String Expr | App Expr Expr

data Value a = V a | F (Value a -> Value a)

inter         


        

Answer 1

How about something like the below. I don't have a good idea on how to traverse Value a, but at least I can see it evaluates example n into MuV.

import Data.Maybe

type Var = String
type MuVar = String

data Expr = Var Var
          | Lam Var Expr
          | App Expr Expr
          | Mu MuVar MuVar Expr
          deriving Show

data Value a = ConV a
             | LamV (Value a -> Value a)
             | MuV (Value a -> Value a)

type Env a = [(Var, Value a)]
type MuEnv a = [(MuVar, Value a -> Value a)]

varScopeErr :: Var -> Value a
varScopeErr v = error $ unwords ["Out of scope λ variable:", show v]

appErr :: Value a
appErr = error "Trying to apply a non-lambda"

muVarScopeErr :: MuVar -> (Value a -> Value a)
muVarScopeErr alpha = id

app :: Value a -> Value a -> Value a
app (LamV f) x = f x
app (MuV f) x = MuV $ \y -> f x `app` y
app _ _ = appErr

eval :: Env a -> MuEnv a -> Expr -> Value a
eval env menv (Var v) = fromMaybe (varScopeErr v) $ lookup v env
eval env menv (Lam v e) = LamV $ \x -> eval ((v, x):env) menv e
eval env menv (Mu alpha beta e) = MuV $ \u ->
  let menv' = (alpha, (`app` u)):menv
      wrap = fromMaybe (muVarScopeErr beta) $ lookup beta menv'
  in wrap (eval env menv' e)
eval env menv (App f e) = eval env menv f `app` eval env menv e

example 0 = App (App t (Var "v")) (Var "w")
  where
    t = Lam "x" $ Lam "y" $ Mu "delta" "phi" $ App (Var "x") (Var "y")
example n = App (example (n-1)) (Var ("z_" ++ show n))